= Solution
For a <Newtonian fluid>, write $\boldsymbol\sigma=-pI+2\mu e$, where $e=(\nabla u+\nabla u^T)/2$ is the <rate-of-strain tensor>. The <Stokes equation> without body force gives $\partial_j\sigma_{ij}=0$. Since the <stress tensor> is symmetric, its contraction with the antisymmetric part of the velocity gradient is zero. Thus
$$
\partial_j(u_i\sigma_{ij})=\sigma_{ij}\partial_ju_i=\boldsymbol\sigma:e.
$$
The <divergence theorem> proves the boundary-work identity
$$
\boxed{D=\int_V\boldsymbol\sigma:e\,dV=\int_{\partial V}u\cdot\boldsymbol\sigma n\,dS.}
$$
For an <incompressible flow>, $\operatorname{tr}e=0$, so $D=2\mu\int_V e:e\,dV\geq0$ is precisely the <viscous dissipation>.
Apply this identity to the particle-free volume and then to the fluid outside the added particle. On the rigid particle, the velocity has the form $V+\Omega\times x$. Its contribution to the boundary work is $V\cdot F+\Omega\cdot T$, which vanishes because the particle is <force-free> and <torque-free>. The outer velocity is unchanged. Subtracting the two boundary-work identities therefore gives
$$
\boxed{D'=\int_{\partial V_0}U_0\cdot\boldsymbol\sigma'n\,dS.}
$$
To move this expression to the particle surface, apply the <Lorentz reciprocal theorem> to $(u_0,\sigma_0)$ and $(u',\sigma')$ in the punctured fluid domain. For completeness, its local proof is
$$
\nabla\cdot(u_0\cdot\sigma'-u'\cdot\sigma_0)
=2\mu e_0:e'-2\mu e':e_0=0.
$$
On the outer boundary $u'=0$. Let $N=-n$ be the particle-outward normal, opposite to the fluid-outward normal there. Integrating the preceding divergence identity gives
$$
\boxed{D'=\int_A(u_0\cdot\sigma'-u'\cdot\sigma_0)N\,dS.}
$$
This expression includes the removal of the original fluid inside the particle; that removal must not be counted separately again.
For a <sphere in a uniform straining Stokes flow>, $E$ is symmetric and trace free, and $u_0=Ex$. Symmetry gives zero rigid translation and rotation. Put $H=x\cdot Ex$ and $r=|x|$. Use the <Unscaled Papkovich–Neuber representation> printed on the paper, with the decaying harmonic potentials
$$
\Phi'=C\frac{a^3Ex}{r^3},\qquad \chi'=B\frac{a^5H}{r^5}.
$$
Their harmonicity follows from $x_i/r^3=-\partial_i(1/r)$ and from $H/r^5$ being a trace-free second derivative of $1/r$. Direct differentiation gives
$$
u'=-3C\frac{a^3H x}{r^5}+B a^5\left(\frac{2Ex}{r^5}-\frac{5H x}{r^7}\right).
$$
At $r=a$, the <no-slip boundary condition> requires $u'=-Ex$. Matching the independent $Ex$ and $Hx$ terms fixes $B=-1/2$ and $C=5/6$. The \b[disturbance velocity and pressure] are consequently
$$
\boxed{u'=-\frac{a^5}{r^5}Ex-\frac{5a^3}{2r^5}\left(1-\frac{a^2}{r^2}\right)H x,
\qquad p'=-\frac{5\mu a^3H}{r^5}.}
$$
The pressure follows from $p'=2\mu\nabla\cdot\Phi'$. The disturbance decays, the total velocity vanishes at the sphere, and the harmonic representation verifies the <Stokes equation>. The boundary traction has zero total force by odd symmetry, and zero total torque because $E$ is symmetric, confirming the assumed <force-free> and <torque-free> motion.
At the sphere $u'=-u_0$, so the <extra dissipation due to a rigid inclusion> becomes $\int_Au_0\cdot(\sigma_0+\sigma')N\,dS$. Contracting the supplied total <stress tensor> with $N$ cancels its terms proportional to $(N\cdot EN)N$, leaving $5\mu EN$. Thus
$$
D'=5\mu a\int_A|EN|^2dS.
$$
Using $\int_AN_iN_jdS=4\pi a^2\delta_{ij}/3$ proves the \b[single-sphere dissipation increase]
$$
\boxed{D'=\frac{20\pi}{3}\mu a^3E:E.}
$$
One may perform the comparison in a large bounded vessel with prescribed straining velocity and then take its size to infinity; the given near-sphere approximation makes this limit compatible with the fixed-outer-velocity definition of $D'$.
The sphere number density is $n_s=\phi/(4\pi a^3/3)=3\phi/(4\pi a^3)$. Neglecting <hydrodynamic interactions>, the additional <viscous dissipation> per unit volume is $n_sD'=5\mu\phi E:E$. Adding the particle-free value $2\mu E:E$ gives
$$
2\mu_{\mathrm{eff}}E:E=2\mu E:E+5\mu\phi E:E,\qquad
\boxed{\mu_{\mathrm{eff}}=\mu\left(1+\frac52\phi\right).}
$$
This is the <Einstein viscosity formula for a dilute suspension>, to first order in the particle <volume fraction>.
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